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as trivial as this question probably seems to everyone, I have become stuck on my attempts to solve the following question:

Given the hyperbola $y = 2/x$ and the circle $x^2 + y^2 = r^2$, what are the values of $r$ where the two graphs have $4$ points of intersection?

Now, I can do this problem easily if I simply graph out both equations and then find the closest position of the hyperbola to the origin (In this case it's $(\sqrt{2},\sqrt{2})$. The answer is $r > 2$. But when I try to solve the two equations simultaneously I can't do it. Would anyone be able to show some possible working out to find the answer with this method.

Many thanks,

Etched

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substitute $y=\frac2{x}$ into $x^2+y^2=r^2$

$$x^2+\frac{4}{x^2}=r^2$$

$$x^4-r^2x^2+4=0$$

$$x^2= \frac{r^2\pm\sqrt{r^4-16}}{2}$$

If the discriminant is positive, $x^2$ take $2$ distinct positive values and hence $x$ takes $4$ distinct values.

$$r^4>16=2^4$$

Hence $r>2$ or $r<-2$. If we are told that $r$ is positive (as it represents the radius), then $r>2$.

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  • $\begingroup$ Geez it's so obvious now :P. Thanks so much! $\endgroup$ – Etched Jan 8 '17 at 0:29

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